Problem 2.42 A generator withVg = 300 V andZg = 50 Ω is connected to a loadZL = 75 Ω through a 50-Ω lossless line of lengthl = 0.15λ .

(a) ComputeZin, the input impedance of the line at the generator end.

(b) ComputeIi andVi .

(c) Compute the time-average power delivered to the line,Pin = 12Re[Vi I∗i ].

(d) Compute VL , IL , and the time-average power delivered to the load,PL = 1

2Re[VL I∗L ]. How doesPin compare toPL? Explain.

(e) Compute the time-average power delivered by the generator,Pg, and the time-average power dissipated inZg. Is conservation of power satisfied?

Solution:

Vg Zin Z0~

+

-

+

-

Transmission line

Generator Loadz = -l z = 0

Vg

IiZg

Zin

~

Vi~~

+

-

⇓ l = 0.15 λ

= 50 Ω

50 Ω

75 Ω

Figure P2.42: Circuit for Problem 2.42.

(a)

β l =2πλ

×0.15λ = 54,

Zin = Z0

[ZL + jZ0 tanβ lZ0 + jZL tanβ l

]= 50

[75+ j50tan54

50+ j75tan54

]= (41.25− j16.35) Ω.

(b)

Ii =Vg

Zg +Zin=

30050+(41.25− j16.35)

= 3.24ej10.16 (A),

Vi = IiZin = 3.24ej10.16(41.25− j16.35) = 143.6e− j11.46 (V).

(c)

Pin =12Re[Vi I

∗i ] =

12Re[143.6e− j11.46 ×3.24e− j10.16 ]

=143.6×3.24

2cos(21.62) = 216 (W).

(d)

Γ =ZL −Z0

ZL +Z0=

75−5075+50

= 0.2,

V+0 = Vi

(1

ejβ l +Γe− jβ l

)=

143.6e− j11.46

ej54 +0.2e− j54 = 150e− j54 (V),

VL = V+0 (1+Γ) = 150e− j54(1+0.2) = 180e− j54 (V),

IL =V+

0

Z0(1−Γ) =

150e− j54

50(1−0.2) = 2.4e− j54 (A),

PL =12Re[VL I∗L ] =

12Re[180e− j54 ×2.4ej54 ] = 216 (W).

PL = Pin, which is as expected because the line is lossless; power input to the lineends up in the load.

(e)Power delivered by generator:

Pg =12Re[VgIi ] =

12Re[300×3.24ej10.16 ] = 486cos(10.16) = 478.4 (W).

Power dissipated in Zg:

PZg =12Re[IiVZg] =

12Re[Ii I

∗i Zg] =

12|Ii |2Zg =

12

(3.24)2×50= 262.4 (W).

Note 1:Pg = PZg +Pin = 478.4 W.